Add the following rational expressions. $\dfrac{6z^2}{3z-1}+\dfrac{2z^2}{3z+1}=$
Solution: We can add two rational expressions whose denominators are equal by adding the numerators and keeping the denominator the same. [Does this fit with how we add rational numbers?] When the denominators are not the same, we must manipulate them so that they become the same. In other words, we must find a common denominator. Since the two denominators do not share any common factors, the common denominator is simply the product of these two denominators: $({3z-1})\cdot({3z+1})$. Let's manipulate the expressions to have that denominator: $\begin{aligned} &\phantom{=}\dfrac{6z^2}{{3z-1}}+\dfrac{2z^2}{{3z+1}} \\\\ &=\dfrac{6z^2\cdot({3z+1})}{({3z-1})\cdot({3z+1})}+\dfrac{2z^2\cdot({3z-1})}{({3z+1})\cdot({3z-1})} \end{aligned}$ [Why did we do that?] Now that both denominators are the same, let's add! $\begin{aligned} &\phantom{=}\dfrac{6z^2\cdot(3z+1)}{(3z-1)\cdot(3z+1)}+\dfrac{2z^2\cdot(3z-1)}{(3z+1)\cdot(3z-1)} \\\\ &=\dfrac{6z^2\cdot(3z+1)+2z^2\cdot(3z-1)}{(3z-1)(3z+1)} \\\\ &=\dfrac{18z^3+6z^2+6z^3-2z^2}{(3z-1)(3z+1)} \\\\ &=\dfrac{24z^3+4z^2}{(3z-1)(3z+1)} \end{aligned}$ In conclusion, $\dfrac{6z^2}{3z-1}+\dfrac{2z^2}{3z+1}=\dfrac{24z^3+4z^2}{(3z-1)(3z+1)}$